8 2 Factoring by GCF Warm Up Identifying

8 -2 Factoring by GCF Warm Up Identifying slope and y-intercept. 1. y = x + 4 m = 1; b = 4 2. y = – 3 x m = – 3; b = 0 Compare and contrast the graphs of each pair of equations. 3. y = 2 x + 4 and y = 2 x – 4 same slope, parallel, and different intercepts 4. y = 2 x + 4 and y = – 2 x + 4 same y-intercepts; different slopes Holt Algebra 1

8 -2 Factoring by GCF Warm Up Simplify. 1. 2(w + 1) 2 w + 2 2. 3 x(x 2 – 4) 3 x 3 – 12 x Find the GCF of each pair of monomials. 3. 4 h 2 and 6 h 2 h 4. 13 p and 26 p 5 13 p Holt Algebra 1

8 -2 Factoring by GCF 8 -2 Holt Algebra 11 Factoring by GCF

8 -2 Factoring by GCF Example 1 A: Factoring by Using the GCF Factor each polynomial. Check your answer. 2 x 2 – 4 2 x 2 = 2 x x 4=2 2 Find the GCF. 2 2 x 2 – (2 2) The GCF of 2 x 2 and 4 is 2. Write terms as products using the GCF as a factor. Use the Distributive Property to factor out the GCF. Multiply to check your answer. The product is the original polynomial. 2(x 2 – 2) Check 2(x 2 – 2) 2 x 2 – 4 Holt Algebra 1

8 -2 Factoring by GCF Example 1 B: Factoring by Using the GCF Factor each polynomial. Check your answer. 8 x 3 – 4 x 2 – 16 x 8 x 3 = 2 2 2 x x x Find the GCF. 4 x 2 = 2 2 x x 16 x = 2 2 x The GCF of 8 x 3, 4 x 2, and 16 x is 4 x. 2 2 x = 4 x Write terms as products using the GCF as a factor. 2 x 2(4 x) – x(4 x) – 4(4 x) Use the Distributive Property to 4 x(2 x 2 – x – 4) factor out the GCF. Check 4 x(2 x 2 – x – 4) Multiply to check your answer. The product is the original 8 x 3 – 4 x 2 – 16 x polynomials. Holt Algebra 1

8 -2 Factoring by GCF Example 1 C: Factoring by Using the GCF Factor each polynomial. Check your answer. – 14 x – 12 x 2 – 1(14 x + 12 x 2) 14 x = 2 7 x 12 x 2 = 2 2 3 x x 2 – 1[7(2 x) + 6 x(2 x)] – 1[2 x(7 + 6 x)] – 2 x(7 + 6 x) Holt Algebra 1 Both coefficients are negative. Factor out – 1. Find the GCF. 2 The GCF of 14 x and 12 x x = 2 x is 2 x. Write each term as a product using the GCF. Use the Distributive Property to factor out the GCF.

8 -2 Factoring by GCF Example 2: Application The area of a court for the game squash is 9 x 2 + 6 x m 2. Factor this polynomial to find possible expressions for the dimensions of the squash court. A = 9 x 2 + 6 x = 3 x(3 x) + 2(3 x) = 3 x(3 x + 2) The GCF of 9 x 2 and 6 x is 3 x. Write each term as a product using the GCF as a factor. Use the Distributive Property to factor out the GCF. Possible expressions for the dimensions of the squash court are 3 x m and (3 x + 2) m. Holt Algebra 1

8 -2 Factoring by GCF Example 3: Factoring Out a Common Binomial Factor each expression. A. 5(x + 2) + 3 x(x + 2)(5 + 3 x) The terms have a common binomial factor of (x + 2). Factor out (x + 2). B. – 2 b(b 2 + 1)+ (b 2 + 1) – 2 b(b 2 + 1) + (b 2 + 1) The terms have a common binomial factor of (b 2 + 1). – 2 b(b 2 + 1) + 1(b 2 + 1) = 1(b 2 + 1)(– 2 b + 1) Holt Algebra 1 Factor out (b 2 + 1).

8 -2 Factoring by GCF Example 3: Factoring Out a Common Binomial Factor each expression. C. 4 z(z 2 – 7) + 9(2 z 3 + 1) There are no common – 7) + + 1) factors. The expression cannot be factored. 4 z(z 2 Holt Algebra 1 9(2 z 3

8 -2 Factoring by GCF Example 4 A: Factoring by Grouping Factor each polynomial by grouping. Check your answer. 6 h 4 – 4 h 3 + 12 h – 8 (6 h 4 – 4 h 3) + (12 h – 8) Group terms that have a common number or variable as a factor. 2 h 3(3 h – 2) + 4(3 h – 2) Factor out the GCF of each group. 2 h 3(3 h – 2) + 4(3 h – 2) is another common factor. (3 h – 2)(2 h 3 + 4) Holt Algebra 1 Factor out (3 h – 2).

8 -2 Factoring by GCF Example 4 B: Factoring by Grouping Factor each polynomial by grouping. Check your answer. 5 y 4 – 15 y 3 + y 2 – 3 y (5 y 4 – 15 y 3) + (y 2 – 3 y) Group terms. 5 y 3(y – 3) + y(y – 3) Factor out the GCF of each group. 5 y 3(y – 3) + y(y – 3) is a common factor. (y – 3)(5 y 3 + y) Factor out (y – 3). Holt Algebra 1

8 -2 Factoring by GCF Recognizing opposite binomials can help you factor polynomials. The binomials (5 – x) and (x – 5) are opposites. Notice (5 – x) can be written as – 1(x – 5) = (– 1)(x) + (– 1)(– 5) Distributive Property. = –x + 5 Simplify. =5–x Commutative Property of Addition. So, (5 – x) = – 1(x – 5) Holt Algebra 1

8 -2 Factoring by GCF Example 5: Factoring with Opposites Factor 2 x 3 – 12 x 2 + 18 – 3 x (2 x 3 – 12 x 2) + (18 – 3 x) Group terms. 2 x 2(x – 6) + 3(6 – x) Factor out the GCF of each group. Write (6 – x) as – 1(x – 6). 2 x 2(x – 6) + 3(– 1)(x – 6) 2 x 2(x – 6) – 3(x – 6)(2 x 2 – 3) Holt Algebra 1 Simplify. (x – 6) is a common factor. Factor out (x – 6).

8 -2 Factoring by GCF Lesson Quiz: Part I Factor each polynomial. Check your answer. 1. 16 x + 20 x 3 4 x(4 + 5 x 2) 2. 4 m 4 – 12 m 2 + 8 m 4 m(m 3 – 3 m + 2) Factor each expression. 3. 7 k(k – 3) + 4(k – 3) 4. 3 y(2 y + 3) – 5(2 y + 3) Holt Algebra 1 (k – 3)(7 k + 4) (2 y + 3)(3 y – 5)

8 -2 Factoring by GCF Lesson Quiz: Part II Factor each polynomial by grouping. Check your answer. 5. 2 x 3 + x 2 – 6 x – 3 (2 x + 1)(x 2 – 3) 6. 7 p 4 – 2 p 3 + 63 p – 18 (7 p – 2)(p 3 + 9) 7. A rocket is fired vertically into the air at 40 m/s. The expression – 5 t 2 + 40 t + 20 gives the rocket’s height after t seconds. Factor this expression. – 5(t 2 – 8 t – 4) Holt Algebra 1

8 -2 Factoring by GCF Warm-Up (Pass Back Papers) Factor each polynomial by grouping. Check your answer. 5. 2 x 3 + x 2 – 6 x – 3 (2 x + 1)(x 2 – 3) 6. 7 p 4 – 2 p 3 + 63 p – 18 (7 p – 2)(p 3 + 9) 7. A rocket is fired vertically into the air at 40 m/s. The expression – 5 t 2 + 40 t + 20 gives the rocket’s height after t seconds. Factor this expression. – 5(t 2 – 8 t – 4) Holt Algebra 1

8 -2 Factoring by GCF Warm-Up (Pass Back Papers) Factor each polynomial. Check your answer. 1. 16 x + 20 x 3 4 x(4 + 5 x 2) 2. 4 m 4 – 12 m 2 + 8 m 4 m(m 3 – 3 m + 2) Factor each expression. 3. 7 k(k – 3) + 4(k – 3) 4. 3 y(2 y + 3) – 5(2 y + 3) Holt Algebra 1 (k – 3)(7 k + 4) (2 y + 3)(3 y – 5)